Answer:
The correct answer would be - 25%.
Explanation:
It is given that the Blue flower is dominant over the yellow flower which means Blue is represented by allele B here and allele b for yellow and Bb is a heterozygous case with a blue phenotypic character. Similarly, the Tall plant is dominant over short and represented by T and t respectively.
BbTt is a heterozygous condition and a cross with itself will produce :
Gametes: BT Bt bT and bt
BT Bt bT bt
BT BBTT BBTt BbTT BbTt
Bt BBTt BBtt BbTt Bbtt
bT BbTT BbTt bbTT bbTt
bt BbTt Bbtt bbTt bbtt
Here 4 out of 16 offspring are heterozygous for both traits represented by bold alphabets. Therefore the correct answer is 25%
The type of glial cells are particularly at risk from the disease are the Schwann cells. This disease is caused by a bacterium, Corynebacterium diphtheriae, and is caused when the bacteria releases a toxin, or poison, into a person's body. The Schwann cells produces insulating myelin sheath that covers the axons of many neurons. These cells may suffers immune or toxic attacks with diphtheria infection.
The correct answer is A. Melting
Explanation:
Melting occurs in solids as these change their state to liquid state, usually due to high temperatures. For example, if one cube of ice is exposed to heat it melts. This physical change implies the solid loss its defined shape and particles in it are not as organized as before.
This process is exemplified by the situation of the oranges because the pyramid of oranges had a defined shape and the oranges were arranged in a defined pattern; however, in the end, this shape changes, and the particles (oranges) are no longer organized in a strict pattern. Also, the distance between particles increases, which also happens when solids become liquids.
Answer:
See the answer below
Explanation:
<em>It may seem that the options have been omitted in this question. However, in order to evaluate the claim that a mean from a particular group is statistically higher or lower to a mean from another group, there would be a need to compare the two means. </em>
Mean comparison or testing for a significant difference between means can be carried out using the Student's t-test or one-way Analysis of Variance (ANOVA).
When the means being compared are just for 2 groups as indicated in the illustration, Student's t-test and one-way ANOVA can be used. However, once there are more than 2 group means being compared, one-way ANOVA will only be applicable.
